Assume that $S$ is an outwardly oriented, piecewise-smooth surface with a piecewise-smooth, simple, closed boundary curve $C$ oriented positively with respect to the orientation of $S$. $ \iint_S \left[ (1 - 2y) \hat{k} \right] \cdot dS$ Use Stokes' theorem to rewrite the surface integral as a line integral. Leave out extraneous functions of $y$ and constant coefficients. $ \oint_C (y^2 \hat{\imath} + $ $ \hat{\jmath} + z \hat{k}) \cdot dr$
Explanation: Assume we have a continuously differentiable three-dimensional vector field $F(x, y, z)$, an oriented piecewise-smooth surface $S$, and a piecewise-smooth, simple, closed boundary curve $C$ oriented positively with respect to $S$. Then Stokes' theorem states that we have the equality below: $ \oint_C F \cdot dr = \iint_S \text{curl}(F) \cdot dS$ If $C$ is negatively oriented, the line integral is equal to the negative of the double integral. [What does any of that mean?] When we use Stokes' theorem to translate from surface integrals to line integrals, we know $\text{curl}(F)$ and we want to find $F$. In general, this inverse curl operation is difficult and sometimes impossible to perform, but in this case we only have one value to find. Specifically: $F(x, y, z) = y^2 \hat{\imath} + ( ??? ) \hat{\jmath} + z \hat{k}$ Our task is to find what $y$ -component gives $F$ the special property: $\text{curl}(F) = (1-2y)\hat{k}$ Let's give a name to the $y$ -component of $F$, say $P$. The only constraint we have on $P$ is that it must make $F$ have the correct curl. $\begin{aligned} \text{curl}(F) &= \det \begin{pmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ \\ y^2 & P & z \end{pmatrix} \\ \\ &= -\dfrac{\partial P}{\partial z} \hat{\imath} + 0 \hat{\jmath} + \left( \dfrac{\partial P}{\partial x} - 2y \right) \hat{k} \end{aligned}$ Matching terms, our original constraint now becomes two specific requirements: $\begin{aligned} -\dfrac{\partial P}{\partial z} &= 0 \\ \\ \dfrac{\partial P}{\partial x} &= 1 \end{aligned}$ One solution is $P = x$. There are infinitely many functions purely of $y$ that we could add on to this solution, but the problem asks us to leave out extraneous functions of $y$. Therefore, we can simplify the surface integral into the line integral below: $ \oint_C (y^2 \hat{\imath} + x \hat{\jmath} + z \hat{k}) \cdot dr$